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\title{Solution} %->->->->-> Check hyperref title <-<-<-<-<-
\author[Yong YANG]{Yong YANG, 2019110294}
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	Beijing University of Posts and Telecommunications%
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	设$A = \begin{bmatrix}
		a & 1-a\\
		b & 1-b
	\end{bmatrix}$($0<a,b<1$), 求$\lim\limits_{n\to\infty}A^n$.
	\begin{block}{解.}
		容易计算出:
		\begin{equation*}
			A^n = \begin{bmatrix}
				\frac{b-(a-1) (a-b)^n}{-a+b+1} & -\frac{(a-1) \left((a-b)^n-1\right)}{a-b-1} \\
				\frac{b \left((a-b)^n-1\right)}{a-b-1} & \frac{b (a-b)^n-a+1}{-a+b+1} \\
			\end{bmatrix}.
		\end{equation*}
		取极限, 得到:
		\begin{equation*}
			\lim\limits_{n\to\infty}A^n = \begin{bmatrix}
			\frac{b}{-a+b+1} & \frac{a-1}{a-b-1} \\
			-\frac{b}{a-b-1} & \frac{-a+1}{-a+b+1} \\
			\end{bmatrix}.
		\end{equation*}
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\textcolor{orange}{谢~~谢}
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